Question

A body of mass m is projected from surface of earth with an initial speed equal to twice the escape velocity. Its speed when it just escapes gravitational pull of earth expressed as a percentage of its initial speed is

(1) 50%

(2) 66.67%

(3) 75%

(4) 86.6%

Answer 1

(4) 86.6%

v_esp = Escape velocity from surface of earth

 \(=\sqrt{\frac{2GM}{R}}\)

M = mass of earth; R = radius of earth. 

v_e = initial speed of projection of body = 3 v_esp 

Let v be its speed when it has just escaped gravitational pull of earth. From law of conservation of energy.